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Electromagnetic Induction

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6 months ago

Electromagnetic Induction Class 12th

A square loop of side 20 cm and resistance is moved towards right with a constant speed v₀. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4. What should be the value of v₀ so that a steady current of 2 mA flows in the loop

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alok kumar singh

Contributor-Level 10

$l = \frac{V_{0} B l}{R} = \frac{V_{0} * 5 * 20 * 1 0^{- 2}}{4 + 1}$

$2 * 1 0^{- 3} = V_{0} * 20 * 1 0^{- 2}$

$V_{0} = \frac{2 * 1 0^{- 3}}{20 * 1 0^{- 2}} = \frac{2}{2} * 1 0^{- 2} = 1 * 1 0^{- 2} m / s$

V₀ = 1 cm/s

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6 months ago

Electromagnetic Induction Class 12th

A coil is placed in a magnetic field as shown below: B

A current is induced in the coil because $\vec{B}$ is:

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Vishal Baghel

Contributor-Level 10

Using faraday's law magnetic field should be outward and decreasing with time

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6 months ago

Electromagnetic Induction Class 12th

A small square look offside 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop offside 'b', then the coefficient of mutual inductance between the two loops is:

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Raj Pandey

Contributor-Level 9

$? = M I$

Also

$? = \vec{B} . \vec{A} = B A c o s 0 °$

= 4 {\frac{μ_{0} l}{4 π (\frac{b}{2})} [\frac{1}{\sqrt[]{2}} + \frac{1}{\sqrt[]{2}}]} . a^{2}

? = 4 * \frac{μ_{0} l}{2 π b} \sqrt[]{2} a^{2}

And on comparing $?$ with

$? = M I$

$\frac{μ_{0}}{4 π} 8 \sqrt[]{2} \frac{a^{2}}{b} = M$

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6 months ago

Electromagnetic Induction Class 12th

A constant magnetic field of 1T is applied in the x > 0 region. A metallic circular ring of radius 1m is moving with a constant velocity of 1 m/s along the x-axis. At t = 0, the centre of O of the ring is at x = -1m. What will be the value of the induced emf in the ring at t = 1s? (Assume the velocity of the ring does not change.)

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alok kumar singh

Contributor-Level 10

$ε = B v (2 R) = 1 * 1 * 2 * 1 = 2 V$

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6 months ago

Electromagnetic Induction Class 12th

A bar magnet is passing through a conducting loop of radius R with velocity $υ .$ The radius of the bar magnet is such that it just passes through the loop. The induced e.m.f. in the loop can be represented by the approximate curve:

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Vishal Baghel

Contributor-Level 10

Emf = $\frac{- d ?}{d t}$ $\frac{}{}$

When magnet passes through the centre rogion of loop, no emf is induced and the polarity of emf will be negative while entering the loop.

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6 months ago

Electromagnetic Induction Class 12th

A circular coil of radius 8.0cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s^-1 in a uniform horizontal magnetic field at 3.0 * 10^-2 T. The maximum emf induced in the coil will be________* 10^-2 volt (rounded off to the nearest integer).

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alok kumar singh

Contributor-Level 10

$ε_{m a x} = n B A ω$

= 20 * 3.0 * 10^-2 * 3.14 * (0.08)² * 50

= 60.3 * 10^-2 V

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6 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

As per the given circuit, the value of current through the battery will be__________A.

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Payal Gupta

Contributor-Level 10

$\begin{array}{l} D_{1} \\ D_{3} \end{array}}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I = \frac{v}{R} = \frac{10}{10} = 1 A m p$

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6 months ago

Electromagnetic Induction physics ncert exemplar solutions class 12th chapter two

A metallic conductor of length 1m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5rad s^-1. If the horizontal component of earth's magnetic field is 0.2 * 10^-4T, then emf induced between the two ends of the conductor is:

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Vishal Baghel

Contributor-Level 10

emf induced = $B_{H} \frac{ω l^{2}}{2}$

$\begin{array}{l} = 0.2 * 1 0^{- 4} * 5 * \frac{1^{2}}{2} \\ = 50 μ v \end{array}$

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6 months ago

Electromagnetic Induction physics ncert exemplar solutions class 12th chapter two

Two coils of self inductance L₁ and L₂ are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be:

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Vishal Baghel

Contributor-Level 10

$L_{e q} = L_{1} + L_{2} - 2 M$

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6 months ago

Electromagnetic Induction Physics NCERT Exemplar Solutions Class 12th Chapter Six

Given below two statements : One is labeled as Assertion (A) and other is labeled as Reason (R).

Assertion (A) : Non-polar materials do not have any permanent dipole moment.

Reason (R) : When a non-polar material is placed in an electric field, the centre of the positive charge distribution of it's individual atom or molecule coincides with the centre of the negativne charge distribution .

In the light of above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

In non-polar molecules, centre of +ve charge coincides with centre of –ve charge, Hence, net dipole moment becomes zero.

When non-polar material is placed in external field, centre of charge does not coincide, hence give non-zero moment.

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